Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
  ↳6 QDP
  ↳7 QDP
  ↳8 QDP

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
*(*(x, y), z) → *(x, *(y, z))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

Q DP problem:
The TRS P consists of the following rules:

+1(s(x), s(y)) → +1(x, y)
+1(+(x, y), z) → +1(x, +(y, z))
+1(+(x, y), z) → +1(y, z)
*1(s(x), s(y)) → +1(*(x, y), +(x, y))
*1(s(x), s(y)) → *1(x, y)
*1(s(x), s(y)) → +1(x, y)
*1(*(x, y), z) → *1(x, *(y, z))
*1(*(x, y), z) → *1(y, z)
SUM(cons(x, l)) → +1(x, sum(l))
SUM(cons(x, l)) → SUM(l)
PROD(cons(x, l)) → *1(x, prod(l))
PROD(cons(x, l)) → PROD(l)

The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
*(*(x, y), z) → *(x, *(y, z))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 4 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

Q DP problem:
The TRS P consists of the following rules:

+1(+(x, y), z) → +1(x, +(y, z))
+1(s(x), s(y)) → +1(x, y)
+1(+(x, y), z) → +1(y, z)

The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
*(*(x, y), z) → *(x, *(y, z))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(6) Obligation:

Q DP problem:
The TRS P consists of the following rules:

SUM(cons(x, l)) → SUM(l)

The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
*(*(x, y), z) → *(x, *(y, z))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(7) Obligation:

Q DP problem:
The TRS P consists of the following rules:

*1(*(x, y), z) → *1(x, *(y, z))
*1(s(x), s(y)) → *1(x, y)
*1(*(x, y), z) → *1(y, z)

The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
*(*(x, y), z) → *(x, *(y, z))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(8) Obligation:

Q DP problem:
The TRS P consists of the following rules:

PROD(cons(x, l)) → PROD(l)

The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
*(*(x, y), z) → *(x, *(y, z))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.